## 物理代写|广义相对论代写General relativity代考|PHY475

statistics-lab™ 为您的留学生涯保驾护航 在代写广义相对论General relativity方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写广义相对论General relativity代写方面经验极为丰富，各种代写广义相对论General relativity相关的作业也就用不着说。

## 物理代写|广义相对论代写General relativity代考|Remarks on the algebra of $\rho$-forms

Let us revert to considering 2-forms; and first show that if $\boldsymbol{\omega}$ and $\boldsymbol{\sigma}$ are both 1 -forms
$$\omega=a_i \boldsymbol{\theta}^i, \quad \boldsymbol{\sigma}=b_k \boldsymbol{\theta}^k,$$
then their wedge product is a 2 -form:
\begin{aligned} \omega \wedge \boldsymbol{\sigma} & =a_i b_k \boldsymbol{\theta}^i \wedge \boldsymbol{\theta}^k \ & =1 / 2\left(a_i b_k-a_k b_i\right) \boldsymbol{\theta}^i \wedge \boldsymbol{\theta}^k \ & =c_{i k} \boldsymbol{\theta}^i \wedge \boldsymbol{\theta}^k \end{aligned}
with
$$c_{i k}=-c_{k i}=1 / 2\left(c_{i k}-c_{k i}\right)=c_{[i k]}$$
the coefficients $c_{i k}$ are the components of an antisymmetric $\left(\begin{array}{l}0 \ 2\end{array}\right)$ tensor. It is then clear that $\boldsymbol{\omega} \wedge \boldsymbol{\sigma}$ is a 2 -form, and also that
$$\omega \wedge \sigma=-\sigma \wedge \omega .$$
Analogous relations hold for general wedge products. Let $\boldsymbol{\alpha}$ be a $p$-form and $\boldsymbol{\beta}$ a $q$-form, so that
$$\boldsymbol{\alpha}=a_{k_1 \ldots k_p} \boldsymbol{\theta}^{k_1} \wedge \cdots \wedge \boldsymbol{\theta}^{k_p}=a_{\left[k_1 \ldots k_p\right]} \boldsymbol{\theta}^{k_1} \wedge \cdots \wedge \boldsymbol{\theta}^{k_p}$$
and the coefficients $a_{\left[k_1 \ldots k_p\right]}$ are the components of a totally antisymmetric $\left(\begin{array}{c}0 \ p\end{array}\right)$ tensor. A similar formula holds for $\boldsymbol{\beta}$ and it then follows, by manipulations similar to those which lead to (3.79), that
$$\alpha \wedge \boldsymbol{\beta}=(-1)^{p q} \beta \wedge \boldsymbol{\alpha}$$

## 物理代写|广义相对论代写General relativity代考|A note on orientation

We saw above that the area $A$ of a parallelogram defined by the vectors $\mathbf{v}, \mathbf{w}$ is $\pm\left|\begin{array}{cc}v_x & v_y \ w_x & w_y\end{array}\right|$ (and $A$ is always taken to be positive). For simplicity take $\mathbf{v}$ and $\mathbf{w}$ to be at right angles, and let us assume that in some local Cartesian coordinate system $\mathbf{v}$ is in the $+x$ direction and $\mathbf{w}$ in the $+y$ direction; then
$$A=\left|\begin{array}{cc} v_x & v_y \ w_x & w_y \end{array}\right|=\left|\begin{array}{cc} v_x & 0 \ 0 & w_y \end{array}\right|>0 .$$
If, however, the $x$ and $y$ axes are interchanged (see Fig. 3.8), we find that
$$\left|\begin{array}{cc} v_x & v_y \ w_x & w_y \end{array}\right|=\left|\begin{array}{cc} 0 & v_y \ w_x & 0 \end{array}\right|<0 .$$ If a space has the property that it is possible to define $A>0$ consistently over the whole space, the space is called orientable. Otherwise it is non orientable. In an orientable space the existence of two distinguishable classes $A>0$ and $A<0$ allows a global distinction between right-handed and left-handed coordinate systems over the space, but this will not hold in a non-orientable one. An example of a non-orientable (2-dimensional) space is the Möbius strip, illustrated in Fig. 3.9. A coordinate system is set up at $P$, and it is seen that after transporting it round the band to $Q$ the $x$ and $y$ axes have been interchanged, so a consistent definition of the sign of $A$ over the surface is not possible. The Möbius strip is actually also an example of a fibre bundle. This is seen as follows: a cylinder is made by drawing a rectangle and joining together the edges marked with arrows, so that the arrows are aligned, as in Fig. 3.10(a). Coordinatising the rectangle, this means that the point $(1, y)$ becomes identified with the point $(1, y)$. If, however, one of the arrows on the rectangle is inverted, then joining the edges in such a way that the arrows are still aligned results in a Möbius strip, as in Fig. 3.10(b). This corresponds to the identification of the points $(1, y)$ and $(1, y)$ in the original rectangle. Now compare the rectangles in (a) and (b). Moving from the point $(1, y)$, keeping $y$ constant but with $x$ decreasing, describes a journey on the cylinder where we eventually return to the original point – so that $x$ has completed a circuit and $y$ has remained unchanged. But on the Möbius strip after $x$ has completed a circuit, from $x=1$ to $x=1, y$ has changed. This means it is not possible to define a Cartesian coordinate system over the whole space; the space may be coordinatised by $(x, y)$, but this does not represent a Cartesian product of $x$ and $y$. Calling the $x$ axis the base space and the $y$ axis the fibre, a closed circuit in the base space results in a motion along the fibre. (The reader will recall that it was argued at the beginning of this chapter that space-time is a fibre bundle.)

# 广义相对论代考

## 物理代写|广义相对论代写General relativity代考|Remarks on the algebra of $\rho$-forms

$$\omega=a_i \boldsymbol{\theta}^i, \quad \boldsymbol{\sigma}=b_k \boldsymbol{\theta}^k,$$

\begin{aligned} \omega \wedge \boldsymbol{\sigma} & =a_i b_k \boldsymbol{\theta}^i \wedge \boldsymbol{\theta}^k \ & =1 / 2\left(a_i b_k-a_k b_i\right) \boldsymbol{\theta}^i \wedge \boldsymbol{\theta}^k \ & =c_{i k} \boldsymbol{\theta}^i \wedge \boldsymbol{\theta}^k \end{aligned}

$$c_{i k}=-c_{k i}=1 / 2\left(c_{i k}-c_{k i}\right)=c_{[i k]}$$

$$\omega \wedge \sigma=-\sigma \wedge \omega .$$

$$\boldsymbol{\alpha}=a_{k_1 \ldots k_p} \boldsymbol{\theta}^{k_1} \wedge \cdots \wedge \boldsymbol{\theta}^{k_p}=a_{\left[k_1 \ldots k_p\right]} \boldsymbol{\theta}^{k_1} \wedge \cdots \wedge \boldsymbol{\theta}^{k_p}$$

$$\alpha \wedge \boldsymbol{\beta}=(-1)^{p q} \beta \wedge \boldsymbol{\alpha}$$

## 物理代写|广义相对论代写General relativity代考|A note on orientation

$$A=\left|\begin{array}{cc} v_x & v_y \ w_x & w_y \end{array}\right|=\left|\begin{array}{cc} v_x & 0 \ 0 & w_y \end{array}\right|>0 .$$

$$\left|\begin{array}{cc} v_x & v_y \ w_x & w_y \end{array}\right|=\left|\begin{array}{cc} 0 & v_y \ w_x & 0 \end{array}\right|<0 .$$如果一个空间有可能在整个空间中一致地定义$A>0$，那么这个空间就被称为可定向空间。否则它是不可定向的。在可定向空间中，两个可区分的类$A>0$和$A<0$的存在允许在空间上区分右手和左手坐标系，但这在不可定向空间中不成立。不可定向(2维)空间的一个例子是Möbius条，如图3.9所示。在$P$处建立了一个坐标系统，可以看到，在将其沿带移动到$Q$后，$x$和$y$轴被交换了，因此不可能在表面上对$A$的符号进行一致的定义。Möbius条实际上也是纤维束的一个例子。如图3.10(a)所示，通过绘制矩形并将箭头标记的边缘连接在一起，使箭头对齐，从而形成一个圆柱体。协调矩形，这意味着点$(1, y)$与点$(1, y)$相等。然而，如果矩形上的一个箭头是倒置的，那么以箭头仍然对齐的方式连接边缘会产生Möbius条，如图3.10(b)所示。这对应于原始矩形中点$(1, y)$和$(1, y)$的标识。现在比较(a)和(b)中的矩形。从点$(1, y)$开始移动，保持$y$不变，但$x$减小，描述了在圆柱体上的旅程，我们最终返回到原始点-因此$x$完成了一个回路，$y$保持不变。但是在Möbius条上，当$x$完成一条赛道后，从$x=1$到$x=1, y$的位置发生了变化。这意味着在整个空间上定义笛卡尔坐标系是不可能的;空间可以通过$(x, y)$进行协调，但这并不表示$x$和$y$的笛卡尔积。将$x$轴称为基空间，将$y$轴称为纤维，基空间中的闭合回路导致沿纤维运动。(读者应该还记得，在本章的开头，我们曾讨论过时空是一个纤维束。)

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 物理代写|广义相对论代写General relativity代考|MATHS565

statistics-lab™ 为您的留学生涯保驾护航 在代写广义相对论General relativity方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写广义相对论General relativity代写方面经验极为丰富，各种代写广义相对论General relativity相关的作业也就用不着说。

## 物理代写|广义相对论代写General relativity代考|Tensors

A vector is
$$\mathbf{V}=V^\mu \mathbf{e}_\mu,$$
where $V^\mu$ are the components in the given basis $\mathbf{e}\mu$. A 1-form is $$\omega=\omega_v \theta^v$$ where $\omega_v$ are the components in the basis $\boldsymbol{\theta}^v$, which is dual to the basis $\mathbf{e}\mu$. We can then define a tensor-more precisely, an $\left(\begin{array}{l}r \ s\end{array}\right)$ tensor-as the geometric object that has components in the space which is the Cartesian product of $r$ basis vectors and $s$ basis 1-forms:
$$\mathbf{T}=T_{\lambda \ldots \mu}^{\alpha \ldots \beta} \mathbf{e}\alpha \otimes \cdots \otimes \mathbf{e}\beta \otimes \boldsymbol{\theta}^\lambda \otimes \cdots \otimes \boldsymbol{\theta}^\mu ;$$
$T^{\alpha \ldots \beta} \beta_{\ldots \mu}$ are the components in the given basis. A vector, then, is a $\left(\begin{array}{l}1 \ 0\end{array}\right)$ tensor and a 1 -form simply write
$$\mathbf{T}=T_{\lambda \ldots \mu}^{\alpha \ldots \beta} \mathbf{e}\alpha \ldots \mathbf{e}\beta \boldsymbol{\theta}^\lambda \ldots \boldsymbol{\theta}^\mu .$$
(We should also note that instead of $\boldsymbol{\theta}^\mu$, the 1 -form basis is sometimes written $\mathbf{e}^\mu$.) In a different basis $\mathbf{e}^{\prime}{ }\alpha, \boldsymbol{\theta}^{\prime \lambda}$ etc., the same tensor is written $$\mathbf{T}=T{\lambda \ldots \mu}^{\prime a \ldots \beta} \mathbf{e}\alpha^{\prime} \otimes \cdots \otimes \mathbf{e}\beta^{\prime} \otimes \boldsymbol{\theta}^{\prime \lambda} \otimes \cdots \otimes \boldsymbol{\theta}^{\prime \mu} .$$

## 物理代写|广义相对论代写General relativity代考|Contraction

Consider $T_{\kappa \ldots \lambda}^{\alpha \ldots \beta}$, an $\left(\begin{array}{l}r \ s\end{array}\right)$ tensor, with $r$ upper and $s$ lower indices. It is said to be of rank $r+s$. If one lower index is put equal to one upper one and a summation is performed over all the relevant components, what results is an $\left(\begin{array}{l}r-1 \ s-1\end{array}\right)$ tensor; its rank has been reduced by 2 . So for example $T^{\alpha \ldots \beta}{ }{\kappa \ldots \beta}$ is an $\left(\begin{array}{c}r-1 \ s-1\end{array}\right)$ tensor (since the summation convention means that the summation is carried out without explicit indication). The proof of this is straightforward, but let us consider for simplicity a particular case. $T\mu^{\alpha \beta}$ is a $\left(\begin{array}{l}2 \ 1\end{array}\right)$ tensor. Our claim is that $T_\alpha^{\alpha \beta}$ is a $\left(\begin{array}{l}1 \ 0\end{array}\right)$ tensor-in other words a vector. The proof depends on the transformation formula (3.48). We have, from (3.48) and (3.20)
$$T_\alpha^{\alpha \alpha \beta}=\frac{\partial x^{\prime \alpha}}{\partial x^\kappa} \frac{\partial x^{\prime \beta}}{\partial x^\lambda} \frac{\partial x^\mu}{\partial x^{\prime \alpha}} T_\mu^{\kappa \lambda}=\frac{\partial x^{\prime \beta}}{\partial x^\lambda} T_\mu^{\mu \lambda}$$
which is precisely the transformation law for a vector, Equation (3.7). This proves our claim.

The symmetric or antisymmetric part of a tensor with respect to either its upper or its lower indices may be defined. The convention is to use round brackets for symmetrisation, so that for example, if $\mathbf{T}$ is an $\left(\begin{array}{l}r \ s\end{array}\right)$ tensor then
\begin{aligned} & T^{(\kappa \ldots \lambda)}{ }{\mu \ldots v} \ & \left.=\frac{1}{r !} \text { sum over all permutations of the } r \text { indices } \kappa \ldots \lambda \text { of } T^{\kappa \ldots \lambda}{ }{\mu \ldots v v}\right} \end{aligned}
is symmetric on interchange of any of its upper indices. Square brackets are used for antisymmetrisation, so that for example
\begin{aligned} & T^{\kappa \ldots \lambda}{ }{[\ldots \ldots v]} \ & =\frac{1}{s !}\left{\text { alternating sum over all permutations of the } s \text { indices } \mu \ldots v \text { of } T^{\kappa \ldots \lambda}{ }{\mu \ldots v}\right} . \end{aligned}
To take some simple examples,
\begin{aligned} & T_\mu^{(\kappa \lambda)}=1 / 2\left(T_\mu^{\kappa \lambda}+T_\mu^{\lambda \kappa}\right), \ & T^{[\kappa \lambda]}{ }\mu=1 / 2\left(T\mu^{\kappa \lambda}-T_\mu^{\lambda \kappa}\right), \ & T_{\rho \sigma}^{[\kappa \lambda \mu]}=1 / 6\left(T_{\rho \sigma}^{\kappa \lambda \mu}+T_{\rho \sigma}^{\lambda \mu \kappa}+T_{\rho \sigma}^{\mu \kappa \lambda}-T_{\rho \sigma}^{\kappa \mu \lambda}-T_{\rho \sigma}^{\mu \lambda \kappa}-T_{\rho \sigma}^{\lambda \kappa \mu}\right) . \ & \end{aligned}

# 广义相对论代考

## 物理代写|广义相对论代写General relativity代考|Tensors

$$\mathbf{V}=V^\mu \mathbf{e}\mu,$$ 其中$V^\mu$是给定基中的分量$\mathbf{e}\mu$。1-形式是$$\omega=\omega_v \theta^v$$，其中$\omega_v$是基$\boldsymbol{\theta}^v$中的分量，它是基$\mathbf{e}\mu$的对偶。然后我们可以定义一个张量——更准确地说，一个$\left(\begin{array}{l}r \ s\end{array}\right)$张量——作为一个几何对象，它在空间中有分量，它是$r$基向量和$s$基1形式的笛卡尔积: $$\mathbf{T}=T{\lambda \ldots \mu}^{\alpha \ldots \beta} \mathbf{e}\alpha \otimes \cdots \otimes \mathbf{e}\beta \otimes \boldsymbol{\theta}^\lambda \otimes \cdots \otimes \boldsymbol{\theta}^\mu ;$$
$T^{\alpha \ldots \beta} \beta_{\ldots \mu}$是给定基中的组成部分。一个向量，是一个$\left(\begin{array}{l}1 \ 0\end{array}\right)$张量，是一个1 -形式
$$\mathbf{T}=T_{\lambda \ldots \mu}^{\alpha \ldots \beta} \mathbf{e}\alpha \ldots \mathbf{e}\beta \boldsymbol{\theta}^\lambda \ldots \boldsymbol{\theta}^\mu .$$
(我们还应该注意，1 -基有时写成$\mathbf{e}^\mu$，而不是$\boldsymbol{\theta}^\mu$。)在不同的基$\mathbf{e}^{\prime}{ }\alpha, \boldsymbol{\theta}^{\prime \lambda}$等等，同样的张量被写出来 $$\mathbf{T}=T{\lambda \ldots \mu}^{\prime a \ldots \beta} \mathbf{e}\alpha^{\prime} \otimes \cdots \otimes \mathbf{e}\beta^{\prime} \otimes \boldsymbol{\theta}^{\prime \lambda} \otimes \cdots \otimes \boldsymbol{\theta}^{\prime \mu} .$$

## 物理代写|广义相对论代写General relativity代考|Contraction

$$T_\alpha^{\alpha \alpha \beta}=\frac{\partial x^{\prime \alpha}}{\partial x^\kappa} \frac{\partial x^{\prime \beta}}{\partial x^\lambda} \frac{\partial x^\mu}{\partial x^{\prime \alpha}} T_\mu^{\kappa \lambda}=\frac{\partial x^{\prime \beta}}{\partial x^\lambda} T_\mu^{\mu \lambda}$$

\begin{aligned} & T^{(\kappa \ldots \lambda)}{ }{\mu \ldots v} \ & \left.=\frac{1}{r !} \text { sum over all permutations of the } r \text { indices } \kappa \ldots \lambda \text { of } T^{\kappa \ldots \lambda}{ }{\mu \ldots v v}\right} \end{aligned}

\begin{aligned} & T^{\kappa \ldots \lambda}{ }{[\ldots \ldots v]} \ & =\frac{1}{s !}\left{\text { alternating sum over all permutations of the } s \text { indices } \mu \ldots v \text { of } T^{\kappa \ldots \lambda}{ }{\mu \ldots v}\right} . \end{aligned}

\begin{aligned} & T_\mu^{(\kappa \lambda)}=1 / 2\left(T_\mu^{\kappa \lambda}+T_\mu^{\lambda \kappa}\right), \ & T^{[\kappa \lambda]}{ }\mu=1 / 2\left(T\mu^{\kappa \lambda}-T_\mu^{\lambda \kappa}\right), \ & T_{\rho \sigma}^{[\kappa \lambda \mu]}=1 / 6\left(T_{\rho \sigma}^{\kappa \lambda \mu}+T_{\rho \sigma}^{\lambda \mu \kappa}+T_{\rho \sigma}^{\mu \kappa \lambda}-T_{\rho \sigma}^{\kappa \mu \lambda}-T_{\rho \sigma}^{\mu \lambda \kappa}-T_{\rho \sigma}^{\lambda \kappa \mu}\right) . \ & \end{aligned}

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 物理代写|广义相对论代写General relativity代考|ASTR3740

statistics-lab™ 为您的留学生涯保驾护航 在代写广义相对论General relativity方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写广义相对论General relativity代写方面经验极为丰富，各种代写广义相对论General relativity相关的作业也就用不着说。

## 物理代写|广义相对论代写General relativity代考|Clock synchronisation

Consider events at two nearby points $A$ and $B$. How do we decide if they are simultaneous? How do we define simultaneity? As usual, we use light; we send a light signal from $A$ to $B$ and back to $A$. Referring to Fig. 2.4, the time on $A$ ‘s world-line which is simultaneous with the event at world time coordinate $x^0$ on $B$ ‘s world-line is defined to be half-way between the emission and reception of the light signals, i.e. at (NB: $\Delta x^0$ below is different from that in (2.62))
$$x^0+\frac{\mathrm{d} x^{0(1)}+\mathrm{d} x^{0(2)}}{2}=x^0-\frac{g_{0 i} \mathrm{~d} x^i}{g_{00}} \equiv x^0+\Delta x^0 .$$
In a rotating frame $g_{0 i} \neq 0$, so $\Delta x^0 \neq 0$. Using the above formula simultaneity may be defined – and clocks therefore synchronised – at points along any open line. An attempt to synchronise clocks at all points along a closed line will in general fail, however, since on return to the starting point the difference in world time recorded will be
$$\Delta x^0=-\oint \frac{g_{0 i}}{g_{00}} \mathrm{~d} x^i$$
In a rotating frame of reference this integral will not vanish, so clock synchronisation is not possible: time is not a single valued parameter in such a situation.

The experimental consequences of this were first revealed in the Sagnac effect. Sagnac found that the interference pattern changed when an interferometer was set in uniform rotation. ${ }^4$ Beams of light traverse a closed path in opposite directions, then meet again at the starting point and interfere (we may refer to Fig. 2.4 again). Now arrange for the whole apparatus to rotate. We may derive an expression for the fringe shift using the formulae above. If the axis of rotation is the $z$ axis then the time discrepancy integrated over one circuit is (see (2.55))
\begin{aligned} \Delta t=\frac{1}{c} \Delta x^0 & =\oint \frac{g_{02} \mathrm{~d} x^2}{g_{00}}=\frac{1}{c^2} \oint \frac{\omega r^2 \mathrm{~d} \phi}{1-\left(\omega^2 r^2\right) / c^2} \ & \approx \frac{\omega}{c^2} \int r^2 \mathrm{~d} \phi=\frac{2 \pi \omega r^2}{c^2}=\frac{2 A \omega}{c^2}, \end{aligned}
where $A=\pi r^2$ is the area enclosed by the path. The associated discrepancy in proper time is
$$\Delta \tau=\sqrt{ }-g_{00} \Delta t \approx \Delta t$$
to leading order. This results in an optical path length
$$l+\Delta l=2 \pi r+c \Delta t=L+\frac{2 A \omega}{c},$$
where $L$ is the ‘undisturbed’ path length. For the light beam travelling in the opposite direction the optical path length is
$$l-\Delta l=L-\frac{2 A \omega}{c} .$$

## 物理代写|广义相对论代写General relativity代考|Inertia: Newton versus Mach

Newton’s first law of motion states that a body subject to no forces remains at rest or continues to move in a straight line with constant speed. (Note that by virtue of Newton’s Principle of Relativity (or of course Special Relativity), these situations are equivalent.) Let us rephrase this by saying ‘remains at rest or continues to move in a straight line with constant speed with respect to $X$ ‘. Then what is $\mathrm{X}$ ? Newton replied ‘Absolute space’ and demonstrated the existence of absolute space with his famous bucket experiment. A bucket of water is held at rest, but hanging by a highly coiled (twisted) rope. The water surface is, of course, flat. The bucket is then released so that the rope begins to uncoil, and the bucket starts to turn. After some time the water also starts to rotate, as it begins to partake of the motion of the bucket. This makes the surface of the water concave, because of the ‘centrifugal’ force on the water. Eventually the rope becomes untwisted and the bucket stops turning; the water, however, is still rotating and has a curved surface. At the beginning of this experiment there is no relative motion between the bucket and the water, and the water surface is flat. Later on, however, when both the water and the bucket are turning, there is also no relative motion, but the surface of the water is curved. The centrifugal force felt by the water is not due to its motion relative to the bucket; it must be caused by its motion relative to absolute space. Inertia results from acceleration or rotation relative to absolute space.

The Austrian physicist, mathematician and philosopher Ernst Mach (1838-1916), however, took a different view, one we should now describe as ‘positivist’; space is not ‘real’, only matter is real. Space is simply an abstraction taken from the set of distance relations between material objects on this view. X cannot be ‘absolute space’, it must be matter – what is more, matter on the cosmological scale. ‘When … we say that a body preserves unchanged its direction and velocity in space, our assertion is nothing more or less than abbreviated reference to the entire universe ‘, 6 by which he meant, in effect, heavenly bodies at large distances, commonly referred to as the ‘fixed stars’. These are thought of as defining a rigid system, while the motion of nearby stars averages out to zero. Our knowledge of the Universe is, of course, more detailed and more sophisticated than that obtaining in Mach’s day. In particular we know that the distribution of matter in the Universe is, to a very good approximation, homogeneous, so that we are not ‘at the centre’ and the ‘fixed stars’ are not ‘near the edge’ of the Universe; moreover, the whole distribution of matter is expanding. Nevertheless, despite our more sophisticated perspective, we may still entertain Mach’s original, and highly interesting, suggestion by identifying $\mathrm{X}$ with an average distribution of masses in the Universe.

# 广义相对论代考

## 物理代写|广义相对论代写General relativity代考|Clock synchronisation

$$x^0+\frac{\mathrm{d} x^{0(1)}+\mathrm{d} x^{0(2)}}{2}=x^0-\frac{g_{0 i} \mathrm{~d} x^i}{g_{00}} \equiv x^0+\Delta x^0 .$$

$$\Delta x^0=-\oint \frac{g_{0 i}}{g_{00}} \mathrm{~d} x^i$$

\begin{aligned} \Delta t=\frac{1}{c} \Delta x^0 & =\oint \frac{g_{02} \mathrm{~d} x^2}{g_{00}}=\frac{1}{c^2} \oint \frac{\omega r^2 \mathrm{~d} \phi}{1-\left(\omega^2 r^2\right) / c^2} \ & \approx \frac{\omega}{c^2} \int r^2 \mathrm{~d} \phi=\frac{2 \pi \omega r^2}{c^2}=\frac{2 A \omega}{c^2}, \end{aligned}

$$\Delta \tau=\sqrt{ }-g_{00} \Delta t \approx \Delta t$$

$$l+\Delta l=2 \pi r+c \Delta t=L+\frac{2 A \omega}{c},$$

$$l-\Delta l=L-\frac{2 A \omega}{c} .$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 物理代写|广义相对论代写General relativity代考|A remark on inertial mass

statistics-lab™ 为您的留学生涯保驾护航 在代写广义相对论General relativity方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写广义相对论General relativity代写方面经验极为丰富，各种代写广义相对论General relativity相关的作业也就用不着说。

## 物理代写|广义相对论代写General relativity代考|A remark on inertial mass

The Equivalence Principle states the equality of gravitational and inertial mass, as we have just seen above. It is worthwhile, however, making the following remark. The inertial mass of a particle refers to its mass (deduced, for example, from its behaviour analysed according to Newton’s laws) when it undergoes non-uniform, or non-inertial, motion. There are, however, two different types of such motion; it may for instance be acceleration in a straight line, or circular motion with constant speed. In the first case the magnitude of the velocity vector changes but its direction remains constant, while in the second case the magnitude is constant but the direction changes. In each of these cases the motion is non-inertial, but there is a conceptual distinction to be made. To be precise we should observe this distinction and denote the two types of mass $m_{\mathrm{i}, \text { acc }}$ and $m_{\mathrm{i}, \text { rot }}$. We believe, without, as far as I know, proper evidence, that they are equal
$$m_{\mathrm{i}, \mathrm{acc}}=m_{\mathrm{i}, \mathrm{rot}}$$
The interesting thing is that Einstein’s formulation of the Equivalence Principle referred to inertial mass measured in an accelerating frame, $m_{\mathrm{i}, \mathrm{acc}}$, whereas the Eötvös experiment, described above, establishes the equality (to within the stated bounds) of $m_{\mathrm{i}, \text { rot }}$ and the gravitational mass. The question is: can an experiment be devised to test the equality of $m_{\mathrm{i}, \text { acc }}$ and $m_{\mathrm{g}}$ ? Or even to test (1.19)?

## 物理代写|广义相对论代写General relativity代考|Tidal forces

The Principle of Equivalence is a local principle. To see this, consider the Einstein box in the gravitational field of the Earth, as in Fig. 1.3. If the box descends over a large distance towards the centre of the Earth, it is clear that two test bodies in the box will approach one another, so over this extended journey it is clear that they are in a genuine gravitational field, and not in an accelerating frame (in which they would stay the same distance apart). In other words, the Equivalence Principle has broken down. We conclude that this principle is only valid as a local principle. Over small distances a gravitational field is equivalent to an acceleration, but over larger distances this equivalence breaks down. The effect is known as a tidal effect, and ultimately is due to the curvature produced by a real gravitational field.
Another way of stating the situation is to note that an object in free fall is in an inertial frame. The effect of the gravitational field has been cancelled by the acceleration of the elevator (the ‘acceleration due to gravity’). The accelerations required to annul the gravitational fields of the two test bodies, however, are slightly different, because they are directed along the radius vectors. So the inertial frames of the two bodies differ slightly. The frames are ‘locally inertial’. The Equivalence Principle treats a gravitational field at a single point as equivalent to an acceleration, but it is clear that no gravitational fields encountered in nature give rise to a uniform acceleration. Most real gravitational fields are produced by more or less spherical objects like the Earth, so the equivalence in question is only a local one.

We may find an expression for the tidal forces which result from this non-locality. Figure 1.4 shows the forces exerted on the two test bodies – call them $\mathrm{A}$ and $\mathrm{B}$ – in the gravitational field of a body at $\mathrm{O}$. They both experience a force towards $\mathrm{O}$ of magnitude where $m$ is the mass of $\mathrm{A}$ and $\mathrm{B}, M$ is the mass of the Earth and $r$ the distance of $\mathrm{A}$ and $\mathrm{B}$ from its centre. In addition, let the distance between $\mathrm{A}$ and $\mathrm{B}$ be $x$. Consider the frame in which $\mathrm{A}$ is at rest. This frame is realised by applying a force equal and opposite to $F_{\mathrm{A}}$, to both $\mathrm{A}$ and $\mathrm{B}$, as shown in Fig. 1.4. In this frame, B experiences a force $F$, directed towards A, which is the vector sum of $F_{\mathrm{B}}$ and $F_{\mathrm{A}}$ :
$$F=2 F_{\mathrm{A}} \sin \alpha=2 F_{\mathrm{A}} \cdot \frac{x}{2 r}=\frac{m M G}{r^3} x .$$
A then observes B to be accelerating towards him with an acceleration given by $F=m \mathrm{~d}^2 x /$ $\mathrm{d} t^2$, i.e.
$$\frac{\mathrm{d}^2 x}{\mathrm{~d} t^2}=-\frac{M G}{r^3} x$$
The $1 / r^3$ behaviour is characteristic of tidal forces.

# 广义相对论代考

## 物理代写|广义相对论代写General relativity代考|A remark on inertial mass

$$m_{\mathrm{i}, \mathrm{acc}}=m_{\mathrm{i}, \mathrm{rot}}$$

## 物理代写|广义相对论代写General relativity代考|Tidal forces

$$F=2 F_{\mathrm{A}} \sin \alpha=2 F_{\mathrm{A}} \cdot \frac{x}{2 r}=\frac{m M G}{r^3} x .$$

$$\frac{\mathrm{d}^2 x}{\mathrm{~d} t^2}=-\frac{M G}{r^3} x$$
The $1 / r^3$ 潮汐力的特性。

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 物理代写|广义相对论代写General relativity代考|Fluids

statistics-lab™ 为您的留学生涯保驾护航 在代写广义相对论General relativity方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写广义相对论General relativity代写方面经验极为丰富，各种代写广义相对论General relativity相关的作业也就用不着说。

## 物理代写|广义相对论代写General relativity代考|Fluids

In order to solve Einstein’s equation (5.9), the energy-momentum tensor $T_{\mu \nu}$ is needed. Also, even though gravity is weak, GR is required. For example, if the universe had a uniform mass density $\bar{\rho}$, the quantity $M / r=4 \pi \bar{\rho} r^2 / 3>1$, at some $r$. In order to make headway, the many bodies making up the universe are subject to a simplifying assumption, namely that they constitute a perfect fluid. Then an momentum-energy tensor, that makes testable predictions, can be obtained.

In the SR frame $\mathrm{O}$ where all the particles are at rest the fluid is called dust. In this frame, the number density of dust particles is $N / V=n \mathrm{~m}^{-3}$. In another SR frame $\mathrm{O}^{\prime}$, where the particles are moving with velocity $\vec{v}^{\prime}$, the volume element is contracted by a factor $1 / \gamma=\left(1-\left|\vec{v}^{\prime}\right|^2\right)^{1 / 2}$. In $\mathrm{O}^{\prime}$ $n^{\prime}=\gamma n$.

The flux of particles across a surface is the number crossing the surface in the direction of the normal to the surface per unit area per unit time. Thus, all particles within a distance $v^{\bar{i}^{\prime}} \Delta t^{\prime}$ of the surface, where $v^{\bar{i}^{\prime}}$ is the speed in the normal direction and within area $\Delta A$ that defines the size of the surface, will cross the surface in time period $\Delta t^{\prime}$,
\begin{aligned} f & =\left(n^{\prime} v^{\bar{i}^{\prime}} \Delta t^{\prime} \Delta A\right) /\left(\Delta t^{\prime} \Delta A\right) \ & =\gamma n v^{i^{\prime}} . \end{aligned}

Note that the vector $N^{\bar{\mu}^{\prime}}=n U^{\bar{\mu}^{\prime}}$ combines both the flux and number density,
\begin{aligned} N^{\overline{0}^{\prime}} & =n U^{\overline{0}^{\prime}}=\gamma n, \ N^{\bar{i}^{\prime}} & =n U^{\bar{i}^{\prime}}=\gamma n v^{\bar{i}^{\prime}}, \ N^{\bar{\mu}^{\prime}} N_{\bar{\mu}^{\prime}} & =\gamma^2 n^2\left(-1+\left(v^{\prime}\right)^2\right)=-n^2 . \end{aligned}

## 物理代写|广义相对论代写General relativity代考|Robertson–Walker Einstein Dynamics

The nonzero C symbols, Ricci tensor and scalar, and Einstein tensor for the Robertson-Walker metric were evaluated in Problems 4.7 and 4.8. The following results were obtained for the $\mathrm{C}$ symbols,
\begin{aligned} \Gamma_{i i}^0 & =g_{i i} \frac{1}{Q} \frac{d Q}{d t}, \quad \Gamma_{0 i}^i=\frac{1}{Q} \frac{d Q}{d t}, \ \Gamma_{r r}^r & =\frac{k r}{1-k r^2}, \sin ^2 \theta \Gamma_{\theta \theta}^r=-r\left(1-k r^2\right) \sin ^2 \theta=\Gamma_{\phi \phi}^r, \ \Gamma_{\theta r}^\theta & =r^{-1}, \Gamma_{\phi \phi}^\theta=-\sin \theta \cos \theta, \quad \Gamma_{\phi r}^\phi=r^{-1}, \Gamma_{\phi \theta}^\phi=\cot \theta . \end{aligned}
These allowed the calculation of the Ricci tensor and scalar,
\begin{aligned} R_{00} & ==-3 \frac{1}{Q} \frac{d^2 Q}{d t^2}=R^{00}, \ R_{i i} & =g_{i i} \frac{1}{Q^2}\left[2\left(k+\left(\frac{d Q}{d t}\right)^2\right)+Q \frac{d^2 Q}{d t^2}\right]=g_{i i}^2 R^{i i}=g_{i i} R^{i i} / g^{i i}, \ R & =g^{\mu \nu} R_{\mu \nu}=g^{\mu \mu} R_{\mu \mu}=6 \frac{1}{Q^2}\left[Q \frac{d^2 Q}{d t^2}+\left(\frac{d Q}{d t}\right)^2+k\right], \end{aligned}
and the Einstein tensor elements,
\begin{aligned} G_{\mu \nu} & =R_{\mu \nu}-g_{\mu \nu} R / 2, \quad G^{\mu \nu}=R^{\mu \nu}-g^{\mu \nu} R / 2, \ G_{00} & =3 \frac{1}{Q^2}\left[\left(\frac{d Q}{d t}\right)^2+k\right], \ G_{i i} & =-g_{i i} \frac{1}{Q^2}\left[2 Q \frac{d^2 Q}{d t^2}+\left(\frac{d Q}{d t}\right)^2+k\right] . \end{aligned}

# 广义相对论代考

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 物理代写|广义相对论代写General relativity代考|Parallel Transport

statistics-lab™ 为您的留学生涯保驾护航 在代写广义相对论General relativity方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写广义相对论General relativity代写方面经验极为丰富，各种代写广义相对论General relativity相关的作业也就用不着说。

## 物理代写|广义相对论代写General relativity代考|Parallel Transport

The characterization of curvature starts with the concept of parallel transport. On a flat surface, as in Fig. 4.2, draw an arbitrary closed path $A B C A$. Here, a circle is used, so that at various places along the path, some tangent vectors $\vec{W}$, that can point in all directions, are shown. For parallel transport start at $A$, draw on the surface, a small parallel transport vector $\vec{V}$ in any direction. Proceed to a neighboring point on the path, draw on the surface a small transport vector, as parallel as possible to the $\vec{V}$ previously drawn. On a flat surface, it is possible to draw the vector exactly parallel. When once again at $A$, the identical vectors would be redrawn. In this sense, a flat surface has no intrinsic curvature. A cylinder can be constructed by rolling a flat sheet, and so has no intrinsic curvature.

A sphere cannot be made from a flat sheet. It has intrinsic curvature. One can find at least one path on the sphere’s surface, as in Fig. 4.3, for which the vectors $\vec{V}$, would not repeat. Pick the path $A B C A$, such that $B$ and $C$ are on the equator, and $A$ is at a pole. At $A$, start with a vector $\vec{V}$ on the sphere’s surface, that is tangent to an arc of longitude. As one proceeds to $B$, along the longitude, a new parallel transport vector cannot be drawn on the surface, exactly parallel to $\vec{V}$. The best one can do is draw that vector along the tangent vector. At $B$ that vector is perpendicular to the equator, and remains so as one proceeds to $C$. From there the return to $A$ is again along a longitude. The parallel transport vectors on the surface will be opposite the tangent vectors. Upon reaching $A$, the final parallel transport vector is different from the initial one.

In spacetime, these vectors have four components $V^\mu, W^\mu$. At any point $P$, one can go to a locally inertial frame. In a small enough neighborhood of $P$, as you proceed along the curve specified by affine parameter $q$ and $W^{\bar{\nu}}=\frac{d x^{\bar{\nu}}}{d q}$, the vector $V^{\bar{\mu}}$ is constant. This leads to a tensor equation, that is taken as the frame invariant definition of parallel transport of $V^\mu$ along $W^\nu$
$$0=\left.\frac{d V^{\bar{\mu}}}{d q}\right|P=V{, \bar{\nu}}^{\bar{\nu}} \frac{d x^{\bar{\nu}}}{d q}=W^{\bar{\nu}} V_{, \bar{\nu}}^{\bar{\mu}}=W^{\bar{\nu}} V^{\bar{\mu}} ; \bar{\nu}=W^\nu V^\mu ;_\nu$$

## 物理代写|广义相对论代写General relativity代考|Curvature Tensors

When gravity is present and there are no boundary surfaces, there is no reason to allow just $x^{1,2}$ to vary, so let them be replaced by generalized coordinates, $x^{\gamma, \lambda}$. Then, the above quantity in parentheses is defined as the Riemann curvature tensor,
$$R_{\beta \gamma \lambda}^\mu=\Gamma_{\lambda \beta}^\nu \Gamma_{\gamma \nu}^\mu-\Gamma_{\gamma \beta}^\mu, \lambda-\Gamma_{\gamma \beta}^\nu \Gamma_{\lambda \nu}^\mu+\Gamma_{\lambda \beta}^\mu,\gamma$$ The proof that $R{\beta \gamma \lambda}^\mu$ is a tensor was carried out in Problem 3.9 , where the following results were obtained for vector $V^\mu$ :
$$V^\mu ;\lambda ; V^\mu{ }{; \gamma} ; \lambda=R_{\beta \lambda \gamma}^\mu V^\beta, R_{\beta \lambda \gamma}^\mu=-R_{\beta \gamma \lambda}^\mu$$
Since the covariant derivative of a tensor is a tensor, the left-hand side of Eq. (4.6) is a tensor of rank 3. The right-hand side of the first equality must be a tensor. Since $V^\beta$ is a tensor of rank $1, R_{\beta \lambda \gamma}^\mu$ must be a tensor of rank 4.

This tensor simplifies for rectangular coordinates in a locally inertial frame because the $\mathrm{C}$ symbols, but not their partial derivatives vanish, and
\begin{aligned} g^{\bar{\mu} \bar{\nu}} ;{\bar{\chi}} & =g^{\bar{\mu} \bar{\nu}}, \bar{\chi}=\eta^{\mu \nu}, \bar{\chi}=0 \ \Gamma{\bar{\lambda} \bar{\beta}}^{\bar{\mu}}, \bar{\gamma} & =\left[\left(g^{\bar{\alpha} \bar{\mu}}\left[g_{\bar{\beta} \bar{\alpha}, \bar{\lambda}}+g_{\bar{\lambda} \bar{\alpha}, \bar{\beta}}-g_{\bar{\beta} \bar{\lambda}, \bar{\alpha}}\right]\right), \bar{\gamma}\right] / 2 \ & =\left(g^{\bar{\alpha} \bar{\mu}}, \bar{\gamma}\left[g_{\bar{\beta} \bar{\alpha}, \bar{\lambda}}+g_{\bar{\lambda} \bar{\alpha}, \bar{\beta}}-g_{\bar{\beta} \bar{\lambda}, \bar{\alpha}}\right]+g^{\bar{\alpha} \bar{\mu}}\left[g_{\bar{\beta} \bar{\alpha}, \bar{\lambda}}+g_{\bar{\lambda} \bar{\alpha}, \bar{\beta}}-g_{\bar{\beta} \bar{\lambda}, \bar{\alpha}}\right], \bar{\gamma}\right) / 2 \ & =\left(g^{\bar{\alpha} \bar{\mu}}\left[g_{\bar{\beta} \bar{\alpha}}, \bar{\lambda}+g_{\bar{\lambda} \bar{\alpha}}, \bar{\beta}-g_{\bar{\beta} \bar{\lambda}}, \bar{\alpha}\right], \bar{\gamma}\right) / 2 \ & =g^{\bar{\alpha} \bar{\mu}}\left(g_{\bar{\beta} \bar{\alpha}}, \bar{\lambda}, \bar{\gamma}+g_{\bar{\lambda} \bar{\alpha}, \bar{\beta}, \bar{\gamma}}-g_{\bar{\beta} \bar{\lambda}, \bar{\alpha}, \bar{\gamma}}\right) / 2 \end{aligned}

# 广义相对论代考

## 物理代写|广义相对论代写General relativity代考|Parallel Transport

$$0=\left.\frac{d V^{\bar{\mu}}}{d q}\right|P=V{, \bar{\nu}}^{\bar{\nu}} \frac{d x^{\bar{\nu}}}{d q}=W^{\bar{\nu}} V_{, \bar{\nu}}^{\bar{\mu}}=W^{\bar{\nu}} V^{\bar{\mu}} ; \bar{\nu}=W^\nu V^\mu ;_\nu$$

## 物理代写|广义相对论代写General relativity代考|Curvature Tensors

$$R_{\beta \gamma \lambda}^\mu=\Gamma_{\lambda \beta}^\nu \Gamma_{\gamma \nu}^\mu-\Gamma_{\gamma \beta}^\mu, \lambda-\Gamma_{\gamma \beta}^\nu \Gamma_{\lambda \nu}^\mu+\Gamma_{\lambda \beta}^\mu,\gamma$$在问题3.9中证明了$R{\beta \gamma \lambda}^\mu$是张量，对于向量$V^\mu$得到如下结果:
$$V^\mu ;\lambda ; V^\mu{ }{; \gamma} ; \lambda=R_{\beta \lambda \gamma}^\mu V^\beta, R_{\beta \lambda \gamma}^\mu=-R_{\beta \gamma \lambda}^\mu$$

\begin{aligned} g^{\bar{\mu} \bar{\nu}} ;{\bar{\chi}} & =g^{\bar{\mu} \bar{\nu}}, \bar{\chi}=\eta^{\mu \nu}, \bar{\chi}=0 \ \Gamma{\bar{\lambda} \bar{\beta}}^{\bar{\mu}}, \bar{\gamma} & =\left[\left(g^{\bar{\alpha} \bar{\mu}}\left[g_{\bar{\beta} \bar{\alpha}, \bar{\lambda}}+g_{\bar{\lambda} \bar{\alpha}, \bar{\beta}}-g_{\bar{\beta} \bar{\lambda}, \bar{\alpha}}\right]\right), \bar{\gamma}\right] / 2 \ & =\left(g^{\bar{\alpha} \bar{\mu}}, \bar{\gamma}\left[g_{\bar{\beta} \bar{\alpha}, \bar{\lambda}}+g_{\bar{\lambda} \bar{\alpha}, \bar{\beta}}-g_{\bar{\beta} \bar{\lambda}, \bar{\alpha}}\right]+g^{\bar{\alpha} \bar{\mu}}\left[g_{\bar{\beta} \bar{\alpha}, \bar{\lambda}}+g_{\bar{\lambda} \bar{\alpha}, \bar{\beta}}-g_{\bar{\beta} \bar{\lambda}, \bar{\alpha}}\right], \bar{\gamma}\right) / 2 \ & =\left(g^{\bar{\alpha} \bar{\mu}}\left[g_{\bar{\beta} \bar{\alpha}}, \bar{\lambda}+g_{\bar{\lambda} \bar{\alpha}}, \bar{\beta}-g_{\bar{\beta} \bar{\lambda}}, \bar{\alpha}\right], \bar{\gamma}\right) / 2 \ & =g^{\bar{\alpha} \bar{\mu}}\left(g_{\bar{\beta} \bar{\alpha}}, \bar{\lambda}, \bar{\gamma}+g_{\bar{\lambda} \bar{\alpha}, \bar{\beta}, \bar{\gamma}}-g_{\bar{\beta} \bar{\lambda}, \bar{\alpha}, \bar{\gamma}}\right) / 2 \end{aligned}

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